Evaluate the following limits :
`lim_(x to 0)(log(3+x)-log(3-x))/x`.

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\log(3+x) - \log(3-x)}{x} \]

we can start by recognizing that substituting \(x = 0\) directly into the expression gives us an indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator.

### Step 1: Differentiate the numerator and denominator

The numerator is \( \log(3+x) - \log(3-x) \). We can differentiate it:

1. The derivative of \( \log(3+x) \) is \( \frac{1}{3+x} \).
2. The derivative of \( \log(3-x) \) is \( -\frac{1}{3-x} \) (using the chain rule).

So, the derivative of the numerator is:

\[ \frac{d}{dx}[\log(3+x) - \log(3-x)] = \frac{1}{3+x} + \frac{1}{3-x} \]

The denominator is simply \( x \), and its derivative is \( 1 \).

### Step 2: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\log(3+x) - \log(3-x)}{x} = \lim_{x \to 0} \frac{\frac{1}{3+x} + \frac{1}{3-x}}{1} \]

### Step 3: Evaluate the limit

Now we substitute \( x = 0 \):

\[ = \frac{1}{3+0} + \frac{1}{3-0} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{\log(3+x) - \log(3-x)}{x} = \frac{2}{3} \]

### Final Answer

\[ \frac{2}{3} \]